Second Order Differential Equations

[roam]

Homework Statement

Solve the following initial value problem:

\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = t

y(0) = 1, \ y'(0) = 0

The correct answer must be: y(t)=3e^{-t} + 2 t e^{-t} + t -2

The Attempt at a Solution

Where did they get the term 2te^-t from?

Here's how I've done it so far:

We can find the general solution of the corresponding homogeneous system:

\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = 0

\frac{d^2 e^{st}}{dt^2} + 2 \frac{de^{st}}{dt}+e^{st} = (s^2 + 2s +1) e^{st} =0

(s+1)(s+1) \implies s=-1

So we have a repeated eigenvalue. Solution to the homogeneous equation is

y_h= C_1 e^{-t}

Now we must find a particular solution, I chose y_p=t-2, since we it is a solution of the nonhomogeneous equation. Therefore the solution is

y = y_h + y_p = C₁ e^-t + t - 2

Now I can use the given initial values to work out C₁:

y' = -C₁e^-t + 1

y(0) = C₁ - 2 = 1 ⇒ C₁ = 3

y'(0) = -C₁+1 = 0

I get y = 3e^-t + t - 2, but there must be another coefficient so we can solve two equations in two unknowns. But how do we get the term 2te^-t? Even if I wrote the repeated eigenvalue twice C₁e^-t + C₂e^-t, I still don't get the term 2te^-t. So what's the problem?

Any help is greatly appreciated.

 

[Mark44]

Homework Statement

Solve the following initial value problem:

\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = t

y(0) = 1, \ y'(0) = 0

The correct answer must be: y(t)=3e^{-t} + 2 t e^{-t} + t -2

The Attempt at a Solution

Where did they get the term 2te^-t from?

Here's how I've done it so far:

We can find the general solution of the corresponding homogeneous system:

\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = 0

\frac{d^2 e^{st}}{dt^2} + 2 \frac{de^{st}}{dt}+e^{st} = (s^2 + 2s +1) e^{st} =0

(s+1)(s+1) \implies s=-1

So we have a repeated eigenvalue. Solution to the homogeneous equation is

y_h= C_1 e^{-t}

No, this is only one solution, and you need two. In cases like this where there is a repeated root of the characteristic equation, you get another solution by multiplying by t.

So two solutions of the homogeneous equation are
\{e^{-t}, te^{-t}\}

The solution to the homogeneous equation would be all linear combinations of the above; $y_h= C_1 e^{-t} + C_2t e^{-t}$

Now we must find a particular solution, I chose y_p=t-2, since we it is a solution of the nonhomogeneous equation. Therefore the solution is

y = y_h + y_p = C₁ e^-t + t - 2

Now I can use the given initial values to work out C₁:

y' = -C₁e^-t + 1

y(0) = C₁ - 2 = 1 ⇒ C₁ = 3

y'(0) = -C₁+1 = 0

I get y = 3e^-t + t - 2, but there must be another coefficient so we can solve two equations in two unknowns. But how do we get the term 2te^-t? Even if I wrote the repeated eigenvalue twice C₁e^-t + C₂e^-t, I still don't get the term 2te^-t. So what's the problem?

Any help is greatly appreciated.

A better particular solution to the nonhomogeneous problem would be y_p = A + Bt

 

[roam]

Oh I see, that makes perfect sense to me know. Thank you so much for your help, I really appreciate it.